Math 487 - Foundations of Geometry Day 15 Group Assignment Name: Theorem 2.11:(The Exterior Angle Theorem) Any exterior angle of a triangle is greater in measure than either of its remote interior angles. 1. Fill in reasons that justify each step in the following proof outline. Proof Outline: Let △ABC be given. Let D be a point such that A − C − D, and thus ∠BCD is an exterior angle of △ABC. Statement Reason Let M be a midpoint of segment BC. B − M − C and BM = MC −−→ There is a point E on ray AM such that A − M − E and ME = MA. ME =∼ MA. ∠AMB and ∠EMC are vertical angles. ∠AMB =∼ ∠EMC △AMB =∼ △EMC ∠ABC = ∠ABM =∼ ∠ECM = ∠ECB. m (∠ABC)= m (∠ECB). ←→ E and D are on the same side of BC. ←→ B, M, and E are on the same side of CD. E ∈ int(∠BCD). m (∠BCE)+ m (∠ECD)= m (∠BCD) m (∠BCD) >m (∠BCE)= m (∠ABC) The proof of the case for the other remote interior angle is similar. ✷. Theorem 2.12: Given a line and a point not on the line, there exists a unique line perpendicular to the given line through the given point. Theorem 2.13: Two lines perpendicular to the same line are parallel. 2. Prove Theorem 2.13 [Hint: Use proof by contradiction and the Exterior Angle Theorem]. Theorem 2.14: There exist at least two lines that are parallel to each other. Theorem 2.15: If there is a transversal to two distinct lines with alternate interior angles congruent, then the two lines are parallel. ←→ ←→ ←→ Proof Outline: Assume the lines AB, DE, and BE are distinct with A − B − C, D − E − F , and G − B − E − H. ←→ Further, assume that A and D are on the same side of line BE. We will prove the contrapositive statement; therefore, ←→ ←→ assume line AB and line DE are not parallel. ←→ ←→ • There exists a point P at the intersection of line AB and line DE. ←→ • Without loss of generality, assume P is on the same side of line BE as the point A. • Consider △PBE • ∠BEF and ∠CBE are exterior angles of △PBE • m(∠PBE) < m(∠BEF ) and m(∠PEB) < m(∠CBE), i.e. m(∠ABE) < m(∠BEF ) and m(∠DEB) < m(∠CBE). • ∠ABE =6∼ ∠BEF and ∠DEB =6∼ ∠CBE ✷ Definitions: • A Saccheri quadrilateral is a quadrilateral ABCD where ∠BAD and ∠ABC are right angles and AD =∼ BC. The segment AB is called the base, and the segment CD is called the summit. • A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. • A rectangle is a quadrilateral with four right angles. 3. Draw a diagram representing a Saccheri quadrilateral. Theorem 2.16: The diagonals of a Saccheri quadrilateral are congruent. 4. Prove Theorem 2.16. Theorem 2.17: The summit angles of a Saccheri quadrilateral are congruent. Proof: Let ABCD be a Saccheri quadrilateral with right angles ∠BAD and ∠ABC, and AD =∼ BC. By Theorem 2.16, AC =∼ BD. Since AD =∼ BC, AC =∼ BD, CD =∼ DC, by SSS congruence [which we have not proven yet], we have △ACD =∼ △BDC. Hence ∠ADC =∼ ∠BCD. Therefore, the summit angles of a Saccheri quadrilateral are congruent. ✷. Theorem 2.18: The segment joining the midpoints of the base and summit of a Saccheri quadrilateral is perpendicular to both the base and summit. Note: You should take time to read the proof of Theorem 2.18 in your textbook between now and class on Friday..